[proofplan] The proof has two ingredients. First, the Kähler commutation identities express the formal adjoints of $\partial_E$ and $\bar\partial_E$ through commutators with $\Lambda$. Second, the curvature identity $\nabla_E^2=\partial_E\bar\partial_E+\bar\partial_E\partial_E=\Theta_h(E)$ converts the difference of the two Laplacians into the curvature commutator. The integral identity follows by taking the $L^2$ inner product with $u$ and using the definition of formal adjoint. [/proofplan] [step:Establish the Kähler commutation identities for the Chern connection] We first prove the operator identities \begin{align*} \bar\partial_E^*=-i[\Lambda,\partial_E], \qquad \partial_E^*=i[\Lambda,\bar\partial_E] \end{align*} on compactly supported smooth $E$-valued forms. Fix a point $x_0\in X$. By the local normal-coordinate construction for Kähler metrics, choose a holomorphic coordinate chart $(U;z_1,\dots,z_n)$ centered at $x_0$ such that \begin{align*} \omega_{x_0}=i\sum_{j=1}^n dz_j\wedge d\bar z_j \end{align*} and the first derivatives of the metric coefficients in these coordinates vanish at $x_0$. Starting from any local holomorphic frame of $E$ over $U$, apply a constant unitary change of frame to make the Hermitian matrix $h_{\alpha\bar\beta}(x_0)$ equal to the identity, and then apply a holomorphic linear correction with prescribed first jet to make $\partial h_{\alpha\bar\beta}(x_0)=0$. In this adjusted holomorphic frame $(e_1,\dots,e_r)$, the Chern connection one-form $h^{-1}\partial h$ vanishes at $x_0$. The vanishing of the connection one-form removes the lower-order connection terms at $x_0$, and the vanishing of the first derivatives of the metric coefficients removes the lower-order terms produced by differentiating the variable-coefficient operator $\Lambda$ at $x_0$. For each $1\leq j\leq n$, define the exterior multiplication operators on the local bundle of $E$-valued forms by \begin{align*} \varepsilon_j:\mathcal A^{\bullet,\bullet}(U,E)&\to \mathcal A^{\bullet+1,\bullet}(U,E), &\alpha&\mapsto dz_j\wedge \alpha,\\ \bar\varepsilon_j:\mathcal A^{\bullet,\bullet}(U,E)&\to \mathcal A^{\bullet,\bullet+1}(U,E), &\alpha&\mapsto d\bar z_j\wedge \alpha. \end{align*} Let $\iota_j$ and $\bar\iota_j$ denote their pointwise Hermitian adjoints with respect to the Hermitian metric determined by $\omega_{x_0}$ and $h_{x_0}$. With the normalization $\omega_{x_0}=i\sum_j dz_j\wedge d\bar z_j$, the adjoint Lefschetz operator at $x_0$ is \begin{align*} \Lambda=-i\sum_{j=1}^n \bar\iota_j\iota_j. \end{align*} The exterior algebra relations give \begin{align*} [\Lambda,\varepsilon_j]=-i\bar\iota_j, \qquad [\Lambda,\bar\varepsilon_j]=i\iota_j. \end{align*} At $x_0$, writing $\nabla_j$ for covariant differentiation in the $\partial/\partial z_j$ direction and $\nabla_{\bar j}$ for covariant differentiation in the $\partial/\partial\bar z_j$ direction, \begin{align*} \partial_E=\sum_{j=1}^n \varepsilon_j\nabla_j, \qquad \bar\partial_E=\sum_{j=1}^n \bar\varepsilon_j\nabla_{\bar j}. \end{align*} Evaluating the coefficient formulas for the formal adjoint differential operators at $x_0$, the normalizations above give \begin{align*} \partial_E^*=-\sum_{j=1}^n \iota_j\nabla_{\bar j}, \qquad \bar\partial_E^*=-\sum_{j=1}^n \bar\iota_j\nabla_j. \end{align*} Therefore \begin{align*} [\Lambda,\partial_E] &= \sum_{j=1}^n [\Lambda,\varepsilon_j]\nabla_j = -i\sum_{j=1}^n \bar\iota_j\nabla_j = i\bar\partial_E^*, \\ [\Lambda,\bar\partial_E] &= \sum_{j=1}^n [\Lambda,\bar\varepsilon_j]\nabla_{\bar j} = i\sum_{j=1}^n \iota_j\nabla_{\bar j} = -i\partial_E^*. \end{align*} Hence $\bar\partial_E^*=-i[\Lambda,\partial_E]$ and $\partial_E^*=i[\Lambda,\bar\partial_E]$ at $x_0$. Since $x_0$ was arbitrary and both sides are globally defined first-order differential operators, the identities hold globally on compactly supported smooth $E$-valued forms. [/step] [step:Convert the difference of Laplacians into the curvature commutator] Set \begin{align*} A:=\partial_E, \qquad B:=\bar\partial_E. \end{align*} All commutators in this step are operator commutators on the common domain $\mathcal A_c^{\bullet,\bullet}(X,E)$ of compactly supported smooth $E$-valued forms. By the identities just proved, \begin{align*} B^*=-i[\Lambda,A], \qquad A^*=i[\Lambda,B]. \end{align*} Using these identities in the definitions of the Laplacians gives \begin{align*} \Delta_{\bar\partial_E} &= BB^*+B^*B = -i\bigl(B[\Lambda,A]+[\Lambda,A]B\bigr), \\ \Delta_{\partial_E} &= AA^*+A^*A = i\bigl(A[\Lambda,B]+[\Lambda,B]A\bigr). \end{align*} Expanding the commutators, \begin{align*} \Delta_{\bar\partial_E} &= -i\bigl(B\Lambda A-BA\Lambda+\Lambda AB-A\Lambda B\bigr), \\ \Delta_{\partial_E} &= i\bigl(A\Lambda B-AB\Lambda+\Lambda BA-B\Lambda A\bigr). \end{align*} Subtracting the second identity from the first, the terms $B\Lambda A$ and $A\Lambda B$ cancel, and we obtain \begin{align*} \Delta_{\bar\partial_E}-\Delta_{\partial_E} &= i(BA+AB)\Lambda-i\Lambda(AB+BA). \end{align*} The curvature of the [Chern connection](/page/Chern%20Connection) is \begin{align*} \nabla_E^2 = (\partial_E+\bar\partial_E)^2 = \partial_E^2+\partial_E\bar\partial_E+\bar\partial_E\partial_E+\bar\partial_E^2. \end{align*} The holomorphic structure is integrable, so $\bar\partial_E^2=0$. The curvature of the Chern connection has type $(1,1)$, so its $(2,0)$ component is zero; since the $(2,0)$ component of $\nabla_E^2$ is $\partial_E^2$, we also have $\partial_E^2=0$. Hence \begin{align*} \nabla_E^2 = \partial_E\bar\partial_E+\bar\partial_E\partial_E = AB+BA. \end{align*} By the statement's definition $\Theta_h(E)=\nabla_E^2$, the operator $AB+BA$ is $\Theta_h(E)$ acting by wedge product of the $\operatorname{End}E$-valued $(1,1)$-form together with the endomorphism action on $E$. Therefore \begin{align*} \Delta_{\bar\partial_E}-\Delta_{\partial_E} &= i\Theta_h(E)\Lambda-\Lambda i\Theta_h(E) = [i\Theta_h(E),\Lambda]. \end{align*} This proves \begin{align*} \Delta_{\bar\partial_E} = \Delta_{\partial_E}+[i\Theta_h(E),\Lambda]. \end{align*} [/step] [step:Take the global inner product on a compact Kähler manifold] Assume now that $X$ is compact, and let $u\in \mathcal A^{\bullet,\bullet}(X,E)$ be smooth. Since $X$ has no boundary and is compact, the formal adjoints $\partial_E^*$ and $\bar\partial_E^*$ are the Hilbert-space adjoints with respect to the $L^2$ inner product induced by $\omega$ and $h$. Hence \begin{align*} (\Delta_{\bar\partial_E}u,u)_{L^2} &= (\bar\partial_E\bar\partial_E^*u,u)_{L^2} + (\bar\partial_E^*\bar\partial_Eu,u)_{L^2} \\ &= \|\bar\partial_E^*u\|_{L^2}^2 + \|\bar\partial_Eu\|_{L^2}^2, \end{align*} and likewise \begin{align*} (\Delta_{\partial_E}u,u)_{L^2} = \|\partial_E^*u\|_{L^2}^2 + \|\partial_Eu\|_{L^2}^2. \end{align*} Taking the $L^2$ inner product of \begin{align*} \Delta_{\bar\partial_E}u = \Delta_{\partial_E}u+[i\Theta_h(E),\Lambda]u \end{align*} with $u$ gives \begin{align*} \|\bar\partial_Eu\|_{L^2}^2+\|\bar\partial_E^*u\|_{L^2}^2 = \|\partial_Eu\|_{L^2}^2+\|\partial_E^*u\|_{L^2}^2 + ([i\Theta_h(E),\Lambda]u,u)_{L^2}. \end{align*} This is the asserted Bochner–Kodaira–Nakano identity. [/step]